Ncert Exercises & Solutions

13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much \({ }_{92}^{235} \mathrm{U}\) did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of \({ }_{92}^{235} \mathrm{U}\) and that this nuclide is consumed only by the fission process.

Hint: One atom of \({ }_{92}^{235} \mathrm{U}\) generates 200 MeV of energy.
Step 1: Find the energy genrated by 1 gm of \({ }_{92}^{235} \mathrm{U}\).
Half-life of the fuel of the fission reactor,

t_{1 / 2} = 5 y e a r s = 5 \times 365 \times 24 \times 60 \times 60 s

We know that in the fission of 1 g of

{}_{92}^{235}U

nucleus, the energy released is equal to 200 MeV.

1 mole, i.e., 235g of

{}_{92}^{235}U

contains

6.023 \times 10^{23}

atoms.

∴ 1 g {}_{92}^{235}U c o n t a i n s \frac{6.023 \times 10^{23}}{235} a t o m s .

The total energy generated per gram of

{}_{92}^{235}U

is calculated as:

E = \frac{6.023 \times 10^{23}}{235} \times 200 M eV / g

= \frac{200 \times 6. 023 \times 10^{23} \times 1.6 \times 10^{- 19} \times 10^{6}}{235} = 8.20 \times 10^{10} J / g

Step 2: Find the total required amount of \({ }_{92}^{235} \mathrm{U}\).
The reactor operates only 80% of the time.

Hence, the amount of

{}_{92}^{235}U

consumed in 5 years by the 1000 MW fission reactor is calculated as:

= \frac{5 \times 80 \times 60 \times 60 \times 365 \times 24 \times 1000 \times 10^{6}}{100 \times 8.20 \times 10^{10}} g

\approx 1538 k g

∴ I n i t i a l a m o u n t o f {}_{92}^{235}U = 2 \times 153 = 3076 k g

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